【发布时间】:2023-03-23 10:02:02
【问题描述】:
我可以在 MongoDB 聚合框架方面为您提供帮助吗?我试图从以下游戏集合中建立一个英超联赛表:
{
"_id" : ObjectId("5b39fec4b5f8df161d259f36"),
"gameWeek" : 1,
"homeTeam" : "Arsenal",
"awayTeam" : "Leicester",
"homeGoals" : 2,
"awayGoals" : 1
}, {
"_id" : ObjectId("5b39ffc2b5f8df161d259f6d"),
"gameWeek" : 2,
"homeTeam" : "Arsenal",
"awayTeam" : "Sunderland",
"homeGoals" : 1,
"awayGoals" : 1
}, {
"_id" : ObjectId("5b39ffe8b5f8df161d259f7f"),
"gameWeek" : 2,
"homeTeam" : "Sunderland",
"awayTeam" : "Manchester United",
"homeGoals" : 1,
"awayGoals" : 1
}, {
"_id" : ObjectId("5b492cbea5aef964f0911cce"),
"gameWeek" : 1,
"homeTeam" : "Manchester United",
"awayTeam" : "Leicester",
"homeGoals" : 0,
"awayGoals" : 1
}
我希望得到以下结果:
{
"_id" : "Arsenal",
"team" : "Arsenal",
"gamesPlayed" : 2,
"goalsFor" : 3,
"goalsAgainst" : 2,
"goalsDifference" : 1,
"gamesWon" : 1,
"gamesDraw" : 1,
"gamesLost" : 0,
"points" : 4
}, {
"_id" : "Leicester",
"team" : "Leicester",
"gamesPlayed" : 2,
"goalsFor" : 2,
"goalsAgainst" : 2,
"goalsDifference" : 0,
"gamesWon" : 1,
"gamesDraw" : 0,
"gamesLost" : 1,
"points" : 3
}, {
"_id" : "Sunderland",
"team" : "Sunderland",
"gamesPlayed" : 2,
"goalsFor" : 2,
"goalsAgainst" : 2,
"goalsDifference" : 0,
"gamesWon" : 0,
"gamesDraw" : 2,
"gamesLost" : 0,
"points" : 2
}, {
"_id" : "Manchester United",
"team" : "Manchester United",
"gamesPlayed" : 2,
"goalsFor" : 1,
"goalsAgainst" : 2,
"goalsDifference" : -1,
"gamesWon" : 0,
"gamesDraw" : 1,
"gamesLost" : 1,
"points" : 1
}
地点:
- gamesPlayed - 玩过的游戏总数,
- goalsFor - 球队的总进球数,
- goalsAgainst - 总进球数,
- goalsDifference - 'goalsFor' 减去 'goalsAgainst'
- points - 每赢一场得 3 分,每场平局得 1 分。
到目前为止,我有以下关于建立团队支持 homeTeam 结果的查询:
db.football_matches.aggregate([
{
$group: {
_id: "$homeTeam",
gamesPlayed : { $sum: NumberInt(1) },
goalsFor: { $sum: "$homeGoals" },
goalsAgainst: { $sum: "$awayGoals" },
gamesWon: { $sum: { $cond: { if: { $gt: [ "$homeGoals", "$awayGoals" ]}, then: NumberInt(1), else: NumberInt(0) } }},
gamesDraw: { $sum: { $cond: { if: { $eq: [ "$homeGoals", "$awayGoals" ]}, then: NumberInt(1), else: NumberInt(0) } }},
gamesLost: { $sum: { $cond: { if: { $lt: [ "$homeGoals", "$awayGoals" ]}, then: NumberInt(1), else: NumberInt(0) } }}
}
}, {
$project: {
team: "$_id" ,
gamesPlayed: "$gamesPlayed",
goalsFor: "$goalsFor",
goalsAgainst: "$goalsAgainst",
goalsDifference: { $subtract: [ "$goalsFor", "$goalsAgainst"] },
gamesWon: "$gamesWon",
gamesDraw: "$gamesDraw",
gamesLost: "$gamesLost",
points: { $add: [ {$multiply: [ "$gamesWon", NumberInt(3)]}, {$multiply: [ "$gamesDraw", NumberInt(1)]} ]}
}
}, {
$sort: { points: -1, goalsDifference: -1 }
}
])
理论上我需要将以下分组结果与另一个类似的 group 语句结合起来,其中将对 awayTeam 字段执行类似的操作:
{
$group: {
_id: "$awayTeam",
gamesPlayed : { $sum: NumberInt(1) },
goalsFor: { $sum: "$awayGoals" },
goalsAgainst: { $sum: "$homeGoals" },
gamesWon: { $sum: { $cond: { if: { $gt: [ "$awayGoals", "$homeGoals" ]}, then: NumberInt(1), else: NumberInt(0) } }},
gamesDraw: { $sum: { $cond: { if: { $eq: [ "$awayGoals", "$homeGoals" ]}, then: NumberInt(1), else: NumberInt(0) } }},
gamesLost: { $sum: { $cond: { if: { $lt: [ "$awayGoals", "$homeGoals" ]}, then: NumberInt(1), else: NumberInt(0) } }}
}
}
我该怎么做?非常感谢。如果之前有人问过类似的问题,请接受我的歉意。
【问题讨论】:
-
这里可以使用
$facet聚合 -
也许正是我需要的。我会尝试。谢谢。
标签: mongodb mongodb-query aggregation-framework